Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

by
Knight, Randall D.

Published by
Pearson

ISBN 10:
0321740904

ISBN 13:
978-0-32174-090-8

Chapter 14 - Oscillations - Exercises and Problems: 8

Answer

$x(t) = (4.0~cm)~cos(8\pi~t+\frac{3\pi}{2})$

Work Step by Step

The general equation for the position of an object in SHM is:
$x(t) = A~cos(2\pi~f~t+\phi_0)$
It is known that:
$A = 4.0 cm$
$f = 4.0 Hz$
At $t = 0$, the object passes through the equilibrium point moving to the right. This shows that the basic cos-curve is shifted an angle of $\frac{3\pi}{2}$ to the left. Therefore, $\phi_0 = \frac{3\pi}{2}$
We can write the function for this motion as:
$x(t) = (4.0~cm)~cos[(2\pi)(4.0)~t+\frac{3\pi}{2}]$
$x(t) = (4.0~cm)~cos(8\pi~t+\frac{3\pi}{2})$